Impedance

Discussion in 'Homework Help' started by ihaveaquestion, May 6, 2009.

  1. ihaveaquestion

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  2. The Electrician

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    Have you studied the use of phasor arithmetic?

    Do you have a calculator that can do complex arithmetic, such as a TI89, or HP50?

    If you don't have such a calculator, how are you doing your complex arithmetic? By hand, or with something like Matlab?
     
  3. ihaveaquestion

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    We didn't learn the phasor stuff...

    And ya, by hand... I think the problem will be made so that the math by hand will be reasonable
     
  4. The Electrician

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    So, can you give an expression for the equivalent impedance of the R-C combination at a frequency of ωo? How would you do that given your studies so far?
     
  5. ihaveaquestion

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  6. The Electrician

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    In the first line where you're doing some algebra for Ztot, you made a mistake in the numerator going from the second fraction to the third.

    When you get to the end, write the final result as two fractions, one with the real part of the numerator over its denominator and another with the imaginary part of the numerator over the same denominator. Show the two fractions with a + sign between them.
     
  7. ihaveaquestion

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  8. The Electrician

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    The left hand part of that final sum is the value R' you're looking for, and the right hand part is the value of 1/jωC'. You can do a little algebra and get the value of C' out of it.

    Now, for the second part, you have to add an impedance Z so that the imaginary part of the sum becomes zero. In other words, add an impedance equal to the negative of the imaginary part of Z1. Then they want to know what is the remaining (real) part?
     
  9. ihaveaquestion

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    Are those correct?

    http://img14.imageshack.us/img14/2144/2353bbm.jpg

    If they are could you explain some logic in this problem to me at how you knew what to do...

    For part a)
    It's asking for an equivalent R'C' SERIES circuit... so this says
    Z = R' + 1/jwC'?

    Somehow I think this relates why you wanted me to separate my term into a+ib form

    For part b)Not really sure about this one... what tipped you to only obtain a real answer? Was it the term 'pure resistance' in the problem?

    Also, my teacher ran through the answer today with 3 minutes of class left... and in her answer she's messing around with jwL which suggests that shes adding an inductance of some type... I dunno
     
  10. The Electrician

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    In part A), you don't have the right stuff in the parentheses in the denominator. When you flipped the fraction over, something went wrong.

    That's exactly it. When you have a resistance R and a reactance X in series, the total impedance is in the form R + JX (the sum of the individual impedances). If they're in parallel, the form is 1/(1/R+1/jX) (the reciprocal of the sum of the reciprocals of the individual impedances), so to find the equivalent series. So you go through the manipulations as you did to find the impedance in the form it needs to be in to represent a pair of series impedances.

    Yes, it was the pure resistance part the tipped me off! And, yes, you need to add an inductive reactance to cancel out the capacitive reactance (this is an example of resonance). Your inductance will have a reactance of jωL. So you need to find L such that jωL = the negative of the imaginary part of Z1.
     
  11. ihaveaquestion

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    Hmm I'm redoing the problem and I noticed that in my impendance equation the j variable is in the numerator, while the R'C' series expression has its j variable in the denominator, is this a problem?

    http://img521.imageshack.us/img521/3320/888n.jpg
     
  12. The Electrician

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    I don't see any asterisk or star in the image linked in your post.

    Just be aware that 1/j = -j and don't make any algebra mistakes and you should be ok.
     
  13. ihaveaquestion

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    http://img10.imageshack.us/img10/5088/09y.jpg

    part a)
    it's not literally asking for a circuit drawing, but the expression that I end up getting, and I don't even necessarily have to solve for R' and C'?

    part b)

    1) making sure - pure resistance = all real, no imaginary?

    2) I determined the value Z, which turns out to be negative. Does a negative impedence make sense? Like a negative value for resistance?

    3) The reason I use the impedance value of the inductor (jwL) is because it asks for a pure resistance value of impedance to add in series, so I need to manipulate the equation because I have j in it by equating it to jwL?
     
  14. ihaveaquestion

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  15. The Electrician

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    You got it just right.

    Don't forget the last question in part b. They want the value of the pure resistance which is just R'.
     
  16. ihaveaquestion

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    I was wondering why she wrote that at the end of her answer in class:

    ZL = R'

    The answer to the first part of part B where it tells me to determine the Z I need to add to make it completely pure (real) is the term
    Z = (jR^2wC) / (1+(RwC)^2)

    When it's asking 'What is the value of this resistance?' I'm not sure if that's a follow-up question to the first part of b), or if it's just asking for the value of R in the circuit
     
  17. The Electrician

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    I think they want you to give them the value of R' in terms of R and C.

    You derived it already. Its R/(1+ω^2*R^2*C^2)
     
  18. ihaveaquestion

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    Just curious because I have a feeling she'll put this type of problem on test -

    for part a) the equivalent R'-C' circuit you get HAS to have the form of R' + 1/jwC' so if one were to end up with a minus sign with j already in the denominator either the problem is flawed or you have the wrong answer from wrong algebra maybe

    for part b) would getting a negative answer make sense.. no right. Because it's asking for the impedance Z that must be added in series, so it has to be positive. I suppose the problem could have variations asking for how much impedance to subtract for a given scenario... in any event manipulating the equation to make it all real is done by either effectively making the element either a capacitance or inductance depending on how you need to cancel your j term (numerator or denominator), correct?
     
  19. The Electrician

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    Part a) Yes, that's probably something to check your answer.

    Part b) If you started with a capacitance, you'll need an inductance to cancel the imaginary part of the impedance, and vice versa.
     
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